4

I have: $$I_{a,b}= \int_b^{ + \infty } \left( \sqrt {\sqrt {x + a} - \sqrt x } - \sqrt {\sqrt x - \sqrt {x - b} } \right)dx$$ with $a>0$ and $b>0$.

I should determine whether this is a convergent or divergent integral. The problem is that I don't know how to start.

Jack D'Aurizio
  • 353,855
Sabaga Bachir
  • 197

1 Answers1

2

We have: $$\sqrt{x+a}-\sqrt{x}=\frac{a}{\sqrt{x}+\sqrt{x+a}},\qquad \sqrt{x}-\sqrt{x-b}=\frac{b}{\sqrt{x}+\sqrt{x-b}}$$ and: $$f_{a,b}(x)=\sqrt{\sqrt{x+a}-\sqrt{x}}-\sqrt{\sqrt{x}-\sqrt{x-b}}=\frac{\frac{a}{\sqrt{x}+\sqrt{x+a}}-\frac{b}{\sqrt{x}+\sqrt{x-b}}}{\frac{\sqrt{a}}{\sqrt{\sqrt{x}+\sqrt{x+a}}}+\frac{\sqrt{b}}{\sqrt{\sqrt{x}+\sqrt{x-b}}}}$$ hence $f_{a,b}(x)$ behaves like $$ x^{1/4}\left(\frac{a}{\sqrt{x}+\sqrt{x+a}}-\frac{b}{\sqrt{x}+\sqrt{x-b}}\right) $$ as $x\to +\infty$. If $a\neq b$, $I_{a,b}$ cannot converge, since $x^{-1/4}\not\in L^1((1,+\infty))$.

If $a=b$ then $f_{a,b}(x)$ behaves like $$ x^{1/4}\left(\frac{1}{\sqrt{x}+\sqrt{x+a}}-\frac{1}{\sqrt{x}+\sqrt{x-a}}\right)=x^{1/4}\left(\frac{\sqrt{x-a}-\sqrt{x+a}}{(\sqrt{x}+\sqrt{x+a})(\sqrt{x}+\sqrt{x-a})}\right)$$ that is $\Theta(x^{1/4-3/2})$ as $x\to +\infty$, hence $I_{a,b}$ converges.

Jack D'Aurizio
  • 353,855
  • 1
    Nice answer, indeed ! What is interesting (assuming no mistake on my side) is that, for $a=b$, the result of the integral seems to be $$-\frac{4}{15} \left(\sqrt{26 \sqrt{2}-14}-2\right) a^{5/4}$$ – Claude Leibovici Sep 01 '14 at 04:58